Download Q.M3 - Tirgul 9

Q.M3 - Tirgul 9
Roee Steiner
Physics Department, Ben Gurion University of the Negev, BeerSheva 84105, Israel
24.12.2014
Contents
1 Gauge transformation
1.1 Classical view . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Quantum view . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Abelian case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
2
3
2 Free particle states: Plane wave versus spherical waves
4
3 Classical scattering
9
1
Gauge transformation
1.1
Classical view
~ and the magnetic field B
~ can be derived from the vector
The electric field E
~
potential A and the electric potential V :
~ = ∇V
E
~ =∇×A
~
B
(1)
(2)
The electric potential and the vector potential are not uniquely determined,
since the electric and the magnetic fields are not affected by the following
changes:
V → V − φ(t)
~→A
~ + ∇λ
A
(3)
(4)
Where λ(x, t) is an arbitrary scalar function and φ(t) is scalar function that
depends on time only. Such a transformation of the potentials is called ” gauge ”.
A special case of gauge is changing the potential V by an addition of a constant.
∗ e-mail:
[email protected]
1
∗
Gauge transformations do not affect the classical motion of the particle since
~,B
~
the equations of motion contain only the derived fields E
1 ~ e~
∂2x
= (eE
− B × ẋ)
∂t2
m
c
(5)
This equation of motion can be derived from the Lagrangian:
L=
1
e ~
mẋ2 + ẋA
− e(V − φ(t))
2
c
(6)
Or, alternatively, from the Hamiltonian:
H=
1.2
1
e~ 2
(~
p − A)
+ e(V − φ(t))
2m
c
(7)
Quantum view
From the Hamiltnian above we can see that the full Schroedinger equation
(namely, with magnetic potential) is:
(i~
~
1
∂
~ + eA )2 ψ(x, t) + V (x)ψ(x, t)
+ φ(t))ψ(x, t) =
(i~∇
∂t
2m
c
(8)
~ is magnetic
where φ is some scalar potential that depends on time only, and A
potential, namely:


i~∂x + eAc x
~
~ + eA = i~∂y + eAy 
i~∇
(9)
c
c
eAz
i~∂z + c
The expatiation value in the quantum mechanics should be the same as in
the classical case. So transformation like equations 3 should not change also the
expatiation value in quantum mechanics.
But unlike the case of classical mechanics (which gauge transformation dont
change any measurable quantity), in quantum mechanics gauge transformation
involves phase transformation of the wave function.
Let see how its work:
~ + eA~ can be written as covariant derivative, namely:
The term i~∇
c
~
~ = i~∇
~ + eA
D
c
(10)
∂
and the term i~ ∂t
+ φ(t):
Dt = i~
∂
+ φ(t)
∂t
(11)
So Schroedinger equation reads:
Dt ψ(x, t) =
1 ~2
D ψ(x, t) + V (x)ψ(x, t)
2m
2
(12)
As we said before the expatiation value should not change, but the wave function
thus involve.
So lets transform the wave function by U , and we will demands that ψ † ψ = ψ 0† ψ 0
where ψ 0 = U ψ. So:
ψ † ψ = ψ 0† ψ 0 = ψ ∗ U † U ψ = ψ † ψ
(13)
So we can conclude that U † U = 1, so U is unitary transformation.
The covariant derivative is gauge invariant, namely:
~ ψ) = U (Dψ)
~
D(U
(14)
Dt (U ψ) = U Dt ψ
(15)
What do I mean gauge invariant? I mean that under transformation the Schrdinger
equation look the same.namely:
1 ~2
D (U ψ(x, t)) − Dt (U ψ(x, t)) + V (x)(U ψ(x, t)) = 0
2m
(16)
Will be:
U
1 ~2
D ψ(x, t) − U Dt ψ(x, t) + U V (x)ψ(x, t) = 0
2m
(17)
1 ~2
D ψ(x, t) − Dt ψ(x, t) + V (x)ψ(x, t)) = 0
2m
(18)
which is:
U(
So the equation of motion save its form.
1.3
Abelian case
In the Abelian case H and U are scalars, so [U,H] = 0, and U can be written
as U = e−iλ(t,x) (because U ∗ U = 1), where λ(t, x) is some scalar function. So:
~
~ ψ) = D(e
~ −iλ(t,x) ψ) = (i~∇
~ + eA )(e−iλ(t,x) ψ)
D(U
c
~
eA
~ + ~∇λ(t,
~
= e−iλ(t,x) (i~∇
x) +
)ψ
c
(19)
We can see that if we want that
~ ψ) = U (Dψ)
~
D(U
(20)
Dt (U ψ) = U Dt ψ
(21)
~ will transform also by
we must have that A
~→A
~ − c~ ∇λ(t,
~
A
x)
e
3
(22)
The same is true for φ
φ → φ − ~∂t λ(t, x)
(23)
We can see that it is transforming like in equations 3
Example
~ = constant, so
We have system with zero magnetic field. This means that A
c~ ~
~
lets say that A = − e ∇λ(t, x) = (0, c0 , 0) in cylinder coordinate. So
∆λ = −
2πe
rc0
c~
(24)
So the wave function get a phase.
2
Free particle states: Plane wave versus spherical waves
The following pages of this section is from Sakurai, which cover the topic in
very good way.
4
6.4
405
Phase Shi fts and Partia l Waves
free-particle Hamiltonian also commutes with L2 and Lz . Thus it is possible to
consider a simultaneous eigenket of Ho,L2 , and Lz . Ignoring spin, such a state is
denoted by I E, l, m} and is often called a spherical-wave state.
More generally, the most general free-particle state can be regarded as a super­
position of I E, l, m} with various E, l, and m in much the same way as the most
general free-particle state can be regarded as a superposition of l k} with different
k, different in both magnitude and direction. Put in another way, a free-particle
state can be analyzed using either the plane-wave basis { l k} } or the spherical-wave
basis { I E, l, m} } .
We now derive the transformation function (kl E, l , m} that connects the plane­
wave basis with the spherical-wave basis. We can also regard this quantity as
the momentum-space wave function for the spherical wave characterized by E, l,
and m. We adopt the normalization convention for the spherical-wave eigenket as
follows:
(E' , l ' , m ' I E, l , m}
= 8ll'8mm'8(E- E').
(6.4. 1 )
In analogy with the position-space wave function, we may guess the angular
dependence:
(ki E, l , m} = gzE(k)Yt (k ),
(6.4.2)
where the function gzE(k) will be considered later. To prove this rigorously, we
proceed as follows. First, consider the momentum eigenket lkZ}-that is, a plane­
wave state whose propagation direction is along the positive z-axis. An important
property of this state is that it has no orbital angular-momentum component in the
z-direction:
Lzl kZ}
= (xpy- YPx) lkx = O, ky = O, kz = k} = 0.
(6.4.3)
Actually this is plausible from classical considerations: The angular-momentum
component must vanish in the direction of propagation because L p = (x x p)
p = 0. Because of (6.4.3)-and since (E ' , l' , m ' lkz} = 0 for m ' =f:. O-we must be
able to expand lkz} as follows:
·
lkZ}
=
LJ
l'
dE ' I E' , l ' , m '
= O} (E ' , l ' , m ' = O lkz} .
·
(6.4.4)
Notice that there is no m ' sum; m ' is always zero. We can obtain the most general
momentum eigenket, with the direction of k specified by() and c/J, from lkZ} by just
applying the appropriate rotation operator as follows [see Figure 3.3 and (3 .6.47)] :
l k}
= :D(a = cp, {J = e, y = O)l kZ} .
(6.4.5)
406
Chapter 6
Scattering Theory
Multiplying this equation by (E, l , m l on the left, we obtain
(E, Z, mlk) =
Lf
l'
x
=
(E' , l ' , m ' = Olkz)
L:j
l'
x
dE ' (E , l , m i:D (a = ¢, {3 = e, y = O)iE' , l ' , m' = 0 )
�6 (a = ¢, f3 = 8, y = 0)
(6.4.6)
dE':D
�� (a = ¢, {3 = e, y = O) (E, l , m = OlkZ) .
8u,8(E - E' ) ( E' , l ' , m ' = OlkZ)
= :D
..[iii!-g1E(k). So we can write, using
Now (E, l , m = O l kZ ) is independent of the orientation of k-that is, independent
of e and ¢-and we may as well call it
(3 .6.5 1),
(kiE , l , m ) = 8ZE (k) Yt
(k).
(6.4.7)
Let us determine gzE(k). First, we note that
(Ho - E ) I E , l , m ) = 0.
(6.4.8)
But we also let Ho - E operate on a momentum eigenbra (kl as follows:
(6.4.9)
Multiplying (6.4.9) with I E, l, m ) on the right, we obtain
(6.4. 1 0)
This means that (kl E, l, m) can be nonvanishing only if E = 1i2 k2 j2m, so we must
be able to write gzE(k ) as
(6 . 4. 1 1 )
6.4
407
Phase S h ifts and Partia l Waves
N
(E',l'm'IE,l,m) I d3k"(E',l',m'lk")(k"IE,l,m)
j k"2dk" j dQ.,,INI20 ( E')
0 ( -E) Y?" (k")Yt(k")
(Ji�-E
2kfl2 ') (Ji�-I k"2dE"
2kfl2 E)
2
d
NI
Q
k"I
- dE"/dk" I
To determine
we go back to our normalization convention (6.4. 1 ). We obtain
=
=
x
x
=
8
Y{;t'* (k")Yt (k")
8
INI2 mk'
Ji2 8(E-E'
E" 1i2k"2 m
)8u'8mm'•
Nk"- 1i rmf
where we have defined
=
j2 to change
integration. Comparing this with (6.4. 1 ), we see that
Therefore, we can finally write
=
(6.4 . 1 2)
E"­
integration into
j
will suffice.
(6.4. 1 3)
hence
2k2 E) Y1 (k).
(kiE,l,m) 1i (1im
lk)
lk) LL I dEIE,l,m)(E,l,mlk)
(�yt*ck:)).
f t IE,Z,m)
=
2
-
m
A
(6.4. 14)
From (6.4. 14) we infer that the plane-wave state
can be expressed as a super­
position of free spherical-wave states with all possible !-values; in particular,
=
l
m
(6.4. 15)
=
l=Om=-1
E=fi2k2j2m
Because the transverse dimension of the plane wave is infinite, we expect that the
plane wave must contain all possible values of impact parameter b (semiclassi­
cally, the impact parameter b:::::: j p ). From this point of view it is no surprise
when analyzed in terms of spherical-wave
that the momentum eigenstates
states, contain all possible values of
We have derived the wave function for
in momentum space. Next, we
consider the corresponding wave function in position space. From wave mechan­
ics, the reader should be familiar with the fact that the wave function for a free
llk),
h
l. IE,l,m)
408
Chapter 6
Scattering Theory
spherical wave isjz(kr) Yt(r), where Jz(kr) is the spherical Bessel function of or­
der [see (3.7.20a) and also Appendix B]. The second solution nt(kr) , although it
satisfies the appropriate differential equation, is inadmissible because it is singular
at the origin. Thus we can write
l
(xiE,l,m) = cziz(kr)YtCn
(6.4. 1 6)
To determine q , all we have to do is compare
LL / dE(xl E ,l, m )(E,l,ml k)
l
h
( h2mk2 ) Y1 (k)
dE czjz(kr)Y1 (r) �
8 E- 2
(xlk) = (2rr)
=
m*
m
=
=
m
A
L1 (214+rr 1 ) Pz(k · r)
A
A
(6.4. 1 7)
h
where we have used the addition theorem
in the last step. Now
(xl k) =
e ik·x /(2rr) 312 can also be written as
(6.4. 18)
which can be proved by using the following integral representation for jz(kr):
jz(kr) =
1
.
2l 1
-
1+1 .
-l
ezkr cose P cos
z( e )
d( os ) .
c
e
(6.4. 19)
Comparing (6.4. 1 7) with (6.4 . 1 8), we have
cz =
iz
h
(6.4.20)
To summarize, we have
h
( h22mk2 ) Y1 (k)
(kiE,l, m ) = �
8 Em
(xiE,l, m ) =
m
jz(kr)Y1
(r).
A
(6.4.21a)
(6.4.21b)
These expressions are extremely useful in developing the partial-wave expansion.
We conclude this section by applying (6.4.21a) to a decay process. Suppose a
parent particle of spin j disintegrates into two spin-zero particles: A (spin j) --+
3
Classical scattering
I took the following pages from the book ”Mechanics from Newton’s laws to
Deterministic Chaos” of ”Florian Scheck”. The aim is to provide Introduction
to the scattering in quantum mechanics.
9
10
Before moving into the solution lates remember the solution of two body system
with central force.
11
12
So
13
14